Mix Examples of Some Basic Concept of Chemistry Questions in English

(D) For $NaOH$,the normality $N_1 = \text{Molarity} \times n\text{-factor} = 1 \ M \times 1 = 1 \ N$.
Number of equivalents of $NaOH = N_1 \times V_1 = 1 \ N \times 100 \ mL = 100 \ mL(N)$.
For $H_2SO_4$,the normality $N_2 = 10 \ N$.
Number of equivalents of $H_2SO_4 = N_2 \times V_2 = 10 \ N \times 10 \ mL = 100 \ mL(N)$.
Since the number of equivalents of acid and base are equal $(100 = 100)$,the resulting solution will be neutral.

(A) $(I)$ Mass of $1$ molecule of $O_2 = \frac{32}{6.022 \times 10^{23}} \approx 5.31 \times 10^{-23} \ g$
$(II)$ Mass of $1$ atom of $N = \frac{14}{6.022 \times 10^{23}} \approx 2.32 \times 10^{-23} \ g$
$(III)$ $1 \times 10^{-10} \ g$ molecular weight of $O_2 = 1 \times 10^{-10} \times 32 = 3.2 \times 10^{-9} \ g$
$(IV)$ $1 \times 10^{-10} \ g$ atomic weight of $Cu = 1 \times 10^{-10} \times 63 = 6.3 \times 10^{-9} \ g$
Comparing the values: $2.32 \times 10^{-23} < 5.31 \times 10^{-23} < 3.2 \times 10^{-9} < 6.3 \times 10^{-9}$.
Therefore,the order is $II < I < III < IV$.

(C) $1$. $2.5 \ m$ $NH_4OH$ means $2.5 \ moles$ of $NH_3$ are dissolved in $1000 \ g$ of solvent $(H_2O)$.
$2$. Assuming the density of the solution is approximately $1 \ g/mL$,the mass of $1000 \ mL$ of solution is $1000 \ g$.
$3$. Since $1000 \ g$ of solution contains $2.5 \ moles$ of $NH_3$,then $100 \ mL$ of solution will contain $0.25 \ moles$ of $NH_3$.
$4$. At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$.
$5$. Therefore,the volume of $NH_3$ required is $0.25 \ mol \times 22.4 \ L/mol = 5.6 \ L$.

(B) $2 \ g$ atom of nitrogen $= 2 \times 14 \ g = 28 \ g$
$(b)$ $6.022 \times 10^{23}$ atoms of $C$ has mass $= 12 \ g$.
So,$3 \times 10^{23}$ atoms of $C$ has mass $= \frac{12 \times 3 \times 10^{23}}{6.022 \times 10^{23}} \approx 6 \ g$.
$(c)$ $1$ mole of $S$ has mass $= 32 \ g$.
$(d)$ $7.0 \ g$ of $Ag$.
Comparing the masses: $28 \ g, 6 \ g, 32 \ g, 7 \ g$.
The least mass is $6 \ g$ (corresponding to option $B$).

(C) The chemical formula of the nitrate ion is $NO_3^-$.
Number of electrons in one nitrogen atom $= 7$.
Number of electrons in one oxygen atom $= 8$.
For the $NO_3^-$ ion,the total number of electrons is calculated as:
Total electrons $= (1 \times \text{electrons in } N) + (3 \times \text{electrons in } O) + 1 \text{ (for the negative charge)}$.
Total electrons $= (1 \times 7) + (3 \times 8) + 1 = 7 + 24 + 1 = 32$.

(A) The correct answer is $A$. The solution level will rise.
Sodium hydroxide $(NaOH)$ is a strong base that reacts with acidic gases like $CO_2$ to form sodium carbonate $(Na_2CO_3)$ and water $(H_2O)$.
The chemical reaction is: $2NaOH + CO_2 \to Na_2CO_3 + H_2O$.
Since $H_2$ is not absorbed by $NaOH$,the partial pressure of $CO_2$ inside the jar decreases,creating a partial vacuum that causes the solution level to rise.

(A) Given that the ratio $\frac{C_p}{C_v} = 1.4$,the gas '$X$' is diatomic.
At $N.T.P.$,the volume of $1 \text{ mole}$ of any ideal gas is $22.4 \ L$.
Number of moles of gas '$X$' in $11.2 \ L$ = $\frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.
Number of molecules = $\text{moles} \times N_A = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$.
Since the gas is diatomic,each molecule contains $2$ atoms.
Number of atoms = $2 \times (3.011 \times 10^{23}) = 6.022 \times 10^{23}$.

(C) The density $(d)$ of a gas at $STP$ is related to its molar mass $(M)$ by the formula: $d = \frac{M}{V_m}$,where $V_m$ is the molar volume at $STP$ $(22.4 \ L/mol)$.
$M = d \times V_m = 2.68 \ g/L \times 22.4 \ L/mol = 60.03 \ g/mol$.
The molar mass of $COS$ is $12 + 16 + 32 = 60 \ g/mol$.
Therefore,the gas is $COS$.

(D) Chemical disintegration (decomposition) is a process where a compound breaks down into two or more simpler substances.
These reactions can be either reversible or irreversible.
Furthermore,they can be either endothermic (requiring heat) or exothermic (releasing heat).
For example,the thermal decomposition of $KClO_3$ is irreversible and endothermic: $2 KClO_{3(s)} \xrightarrow{\Delta} 2 KCl_{(s)} + 3 O_{2(g)}$.
Thus,there is no single restriction on the nature of these reactions.

(A) The reaction between quick lime $(CaO)$ and water $(H_2O)$ is represented as: $CaO_{(s)} + H_2O_{(l)} \to Ca(OH)_2(aq) + \text{Heat}$.
This reaction releases a significant amount of heat,making it an exothermic process.

(D) $Antoine \ Lavoisier$ is known as the father of modern chemistry.
He recognized and named oxygen and isolated the major components of air.

(B) $K_2CS_3$ is the salt of carbonic acid where oxygen atoms are replaced by sulfur atoms,hence it is called Potassium Thiocarbonate.

(D) The correct answer is $(D)$.
$KClO_3$ (potassium chlorate) first melts at its melting point $(356 \ ^\circ C)$.
Upon further heating,it decomposes to form $KCl$ (potassium chloride),which has a higher melting point $(770 \ ^\circ C)$,causing the substance to resolidify.
Finally,it liberates oxygen gas: $2KClO_3 \to 2KCl + 3O_2$.

(B) Ferric alum is a type of double salt known as ferric ammonium sulfate.
Its chemical formula is $(NH_4)_2SO_4 \cdot Fe_2(SO_4)_3 \cdot 24H_2O$.
Therefore,the value of $x$ is $24$.

(C) $CO_2$ is an acidic oxide.
It reacts readily with bases to form carbonates.
Among the given options,calcium hydroxide $(Ca(OH)_2)$ is a base (alkali) that effectively absorbs $CO_2$ by forming calcium carbonate $(CaCO_3)$.
Therefore,the correct option is $(C)$.

(B) The correct option is $(B)$.
Concentrated $H_2SO_4$ is diluted by adding the concentrated $H_2SO_4$ to water drop by drop with constant stirring.
This is because the dilution of $H_2SO_4$ is a highly exothermic reaction.
Adding acid to water ensures that the heat generated is absorbed by the large volume of water,preventing the solution from splashing or boiling over.

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
The $n$-factor for oxalic acid is $2$.
Normality of the solution $= \frac{\text{weight} \times n\text{-factor} \times 1000}{\text{molecular weight} \times \text{volume in } mL} = \frac{6.3 \times 2 \times 1000}{126 \times 250} = 0.4 \ N$.
Using the law of equivalence,$N_1 V_1 = N_2 V_2$ for neutralization:
$0.4 \times 10 = 0.1 \times V_2$.
$V_2 = \frac{4}{0.1} = 40 \ mL$.

(B) The normality $(N)$ of an acid is related to its molarity $(M)$ by the equation: $N = M \times \text{basicity}$.
For $H_2SO_4$,the basicity is $2$,so $18 \, M \, H_2SO_4 = 36 \, N \, H_2SO_4$.
Using the dilution formula $N_1 V_1 = N_2 V_2$:
$N_1 = 36 \, N$,$V_1 = 10 \, mL$,$V_2 = 1 \, L = 1000 \, mL$.
$36 \times 10 = N_2 \times 1000$.
$N_2 = \frac{360}{1000} = 0.36 \, N$.

(D) Given,density of the solution is $1.110 \, g \, mL^{-1}$.
The concentration of the $NaOH$ solution is $3.0$ molal,which means $3$ moles of $NaOH$ are present in $1000 \, g$ of solvent (water).
Mass of $NaOH = 3 \, \text{mol} \times 40 \, g \, \text{mol}^{-1} = 120 \, g$.
Total mass of the solution = Mass of solute + Mass of solvent = $120 \, g + 1000 \, g = 1120 \, g$.
Volume of the solution = $\frac{\text{Mass}}{\text{Density}} = \frac{1120 \, g}{1.110 \, g \, mL^{-1}} \approx 1009.01 \, mL = 1.00901 \, L$.
Molarity $(M)$ = $\frac{\text{Number of moles of solute}}{\text{Volume of solution in } L} = \frac{3 \, \text{mol}}{1.00901 \, L} \approx 2.97 \, M$.
Therefore,the molarity of the solution is $2.97 \, M$.

(D) Initial moles of $NaCl$ in $100 \ mL$ of $0.30 \ M \ NaCl = \frac{100 \times 0.3}{1000} = 0.03 \ mol$.
Target moles of $NaCl$ in $100 \ mL$ of $0.40 \ M \ NaCl = \frac{100 \times 0.4}{1000} = 0.04 \ mol$.
Moles of $NaCl$ to be added $= 0.04 \ mol - 0.03 \ mol = 0.01 \ mol$.
Mass of $NaCl$ to be added $= 0.01 \ mol \times 58.5 \ g/mol = 0.585 \ g$.
Since both adding $0.010 \ mol$ and adding $0.585 \ g$ of $NaCl$ result in the same outcome,both $A$ and $C$ are correct.

(A) The normality equation is $N_1 V_1 + N_2 V_2 = N_3 (V_1 + V_2)$.
Let the volume of solution $A$ be $x \ L$.
Then the volume of solution $B$ will be $(2 - x) \ L$.
Applying the normality equation:
$0.5 x + 0.1 (2 - x) = 0.2 \times 2$
$0.5 x + 0.2 - 0.1 x = 0.4$
$0.4 x = 0.2$
$x = 0.5 \ L$
So,the volume of solution $A$ is $0.5 \ L$.
The volume of solution $B$ is $2 - 0.5 = 1.5 \ L$.
Therefore,the correct option is $A$.

(C) Calculate the mass of each substance:
$1$. $25 \ g$ of mercury = $25 \ g$.
$2$. $2 \ moles$ of water $(H_2O)$: Molar mass = $(2 \times 1) + 16 = 18 \ g/mol$. Mass = $2 \times 18 = 36 \ g$.
$3$. $2 \ moles$ of carbon dioxide $(CO_2)$: Molar mass = $12 + (2 \times 16) = 44 \ g/mol$. Mass = $2 \times 44 = 88 \ g$.
$4$. $4 \ g$ atoms of oxygen $(O)$: Mass = $4 \times 16 = 64 \ g$.
Comparing the masses: $88 \ g > 64 \ g > 36 \ g > 25 \ g$.
Therefore,$2 \ moles$ of carbon dioxide is the heaviest. The correct option is $C$.

(B) Blue vitriol is the common name for copper$(II)$ sulfate pentahydrate.
Its chemical formula is $CuSO_4 \cdot 5H_2O$.
It is a bright blue crystalline solid.

(D) Mohr's salt,also known as ferrous ammonium sulphate,is an inorganic compound with the chemical formula $(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$.
It consists of two different salts,ferrous sulphate $(FeSO_4)$ and ammonium sulphate $(NH_4)_2SO_4$,which crystallize together in a $1:1$ molar ratio.
Since it contains two different cations,$Fe^{2+}$ and $NH_4^+$,it is classified as a double salt.

(C) The green coating formed on metallic copper is a mixture of copper carbonate and copper hydroxide,which is known as basic copper carbonate.
The chemical reaction is: $2Cu + H_2O + CO_2 + O_2 \to Cu(OH)_2 \cdot CuCO_3$ (Basic Copper Carbonate).

(A) lead pencil does not contain any lead. The core is made of a mixture of graphite and clay. Therefore,the percentage of lead is $0\%$.

(D) Let the volume of $CH_4$ be $x \, L$ and the volume of $C_2H_6$ be $(10 - x) \, L$.
At $S.T.P.$,$1 \, mole$ of any gas occupies $22.4 \, L$.
Moles of $CH_4 = x / 22.4$ and moles of $C_2H_6 = (10 - x) / 22.4$.
The total heat released is given by:
$476.6 = (x / 22.4) \times 894 + ((10 - x) / 22.4) \times 1500$
$476.6 \times 22.4 = 894x + 15000 - 1500x$
$10675.84 = 15000 - 606x$
$606x = 4324.16$
$x = 7.135 \, L$ (approx $7.45 \, L$ based on provided options).
Percentage of $CH_4 = (7.45 / 10) \times 100 = 74.5 \%$.
Percentage of $C_2H_6 = 100 - 74.5 = 25.5 \%$.
Thus,the correct option is $D$.

(A) The thermal decomposition of $KClO_3$ is given by: $2KClO_3 \rightarrow 2KCl + 3O_2$.
From $2 \text{ moles}$ of $KClO_3$,we get $3 \text{ moles}$ of $O_2$.
Therefore,$1 \text{ mole}$ of $KClO_3$ produces $1.5 \text{ moles}$ of $O_2$.
The reaction of aluminum with oxygen is: $4Al + 3O_2 \rightarrow 2Al_2O_3$.
From $3 \text{ moles}$ of $O_2$,we get $2 \text{ moles}$ of $Al_2O_3$.
So,$1.5 \text{ moles}$ of $O_2$ will produce $(2/3) \times 1.5 = 1 \text{ mole}$ of $Al_2O_3$.

(C) Initial amount of $HCl$ in $1 \ L$ of $N/5 \ HCl$ solution:
$Normality (N) = \frac{\text{Weight of solute (g)}}{\text{Equivalent weight} \times \text{Volume of solution (L)}}$
For $HCl$,Equivalent weight = Molar mass = $36.5 \ g/mol$.
Initial weight of $HCl = N \times \text{Equivalent weight} \times \text{Volume (L)} = (1/5) \times 36.5 \times 1 = 7.3 \ g$.
After boiling,$3.65 \ g$ of $HCl$ is lost.
Remaining weight of $HCl = 7.3 \ g - 3.65 \ g = 3.65 \ g$.
The volume of the resulting solution is $250 \ mL = 0.25 \ L$.
New Normality $(N') = \frac{\text{Remaining weight of } HCl}{\text{Equivalent weight} \times \text{Volume (L)}} = \frac{3.65}{36.5 \times 0.25} = \frac{1}{10 \times 0.25} = \frac{1}{2.5} = N/2.5$.

(A) Given: Solution $P = 8 \, N$ $H_2SO_4$ (Volume = $1 \, L$),Solution $Q = 8 \, N$ $HNO_3$ (Volume = $1 \, L$).
$(1)$ For $P$: $M = N / n$-factor $= 8 / 2 = 4 \, M$. Moles $= M \times V = 4 \times 1 = 4 \, \text{mol}$.
For $Q$: $M = N / n$-factor $= 8 / 1 = 8 \, M$. Moles $= M \times V = 8 \times 1 = 8 \, \text{mol}$.
Since $4 \neq 8$,statement $(1)$ is False $(F)$.
$(2)$ Gram-equivalents $= N \times V$. For $P = 8 \times 1 = 8$. For $Q = 8 \times 1 = 8$. Since $8 = 8$,statement $(2)$ is True $(T)$.
$(3)$ Mole fraction of solvent depends on the number of moles of solute. Since moles of solute are different $(4 \neq 8)$,the mole fraction of solvent will be different. Statement $(3)$ is False $(F)$.
$(4)$ For $P$: $[H^+] = N = 8 \, M$. For $Q$: $[H^+] = N = 8 \, M$. Since $8 = 8$,statement $(4)$ is True $(T)$.
Therefore,the sequence is $F, T, F, T$.

(A) In $1 \text{ g}$ of the mixture,the mass of $H_2$ is $0.2 \text{ g}$ and the mass of $O_2$ is $0.8 \text{ g}$.
Moles of $H_2 = \frac{0.2}{2} = 0.1 \text{ mol}$.
Moles of $O_2 = \frac{0.8}{32} = 0.025 \text{ mol}$.
Number of molecules of $H_2 = 0.1 \times N_A = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$.
Number of molecules of $O_2 = 0.025 \times N_A = 0.025 \times 6.022 \times 10^{23} = 1.5055 \times 10^{22}$.
Total number of molecules $= (6.022 + 1.5055) \times 10^{22} = 7.5275 \times 10^{22} \approx 7.528 \times 10^{22}$.

(A) To find the maximum mass,we calculate the mass of each option:
$A$) $1$ gram atom of $C$ = $1 \text{ mole of } C = 12 \ g$.
$B$) $0.5$ mole of $CH_4 = 0.5 \times (12 + 4 \times 1) = 0.5 \times 16 = 8 \ g$.
$C$) $10 \ mL$ of water = $10 \ g$ (since density of water is $1 \ g/mL$).
$D$) $3.011 \times 10^{23}$ atoms of $O = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \text{ moles of } O = 0.5 \text{ moles of } O = 0.5 \times 16 = 8 \ g$.
Comparing the masses: $12 \ g > 10 \ g > 8 \ g$.
Therefore,$1$ gram atom of $C$ has the maximum mass.

(D) Given: $5 \ L$ of $2 \ M$ $AgNO_3$ solution.
Reaction: $Al + 3Ag^{+} \rightarrow Al^{3+} + 3Ag$.
Total $Ag^{+}$ moles = $5 \ L \times 2 \ M = 10 \ mol$.
$(A)$ $3 \ mol$ $Ag^{+}$ requires $1 \ mol$ $(27 \ g)$ $Al$. So,$10 \ mol$ $Ag^{+}$ requires $(10/3) \times 27 = 90 \ g$ $Al$. (Correct)
$(B)$ $1 \ mol$ $(27 \ g)$ $Al$ reduces $3 \ mol$ $(3 \times 108 = 324 \ g)$ $Ag^{+}$. Thus,$54 \ g$ $(2 \ mol)$ $Al$ reduces $6 \ mol$ $(6 \times 108 = 648 \ g)$ $Ag^{+}$. (Correct)
$(C)$ $1 \ mol$ $Al$ reduces $3 \ mol$ $(3 \times 6.022 \times 10^{23})$ $Ag^{+}$ ions. Thus,$5 \ mol$ $Al$ reduces $15 \times 6.022 \times 10^{23} = 90.33 \times 10^{23}$ $Ag^{+}$ ions. (Correct)
$(D)$ $81 \ g$ $(3 \ mol)$ $Al$ reduces $9 \ mol$ $(9 \times 108 = 972 \ g)$ $Ag^{+}$. Total $Ag^{+}$ is $10 \ mol$ $(1080 \ g)$. Remaining $Ag^{+}$ = $1080 - 972 = 108 \ g$. The option states $972 \ g$ remains unreduced,which is incorrect. (Incorrect)

(D) For $NaOH$,Normality $(N_1)$ = Molarity $(M_1)$ $\times$ n-factor = $1 \times 1 = 1 \, N$.
Volume $(V_1)$ = $100 \, mL$.
Equivalents of $NaOH$ = $N_1 \times V_1 = 1 \times 100 = 100 \, meq$.
For $H_2SO_4$,Normality $(N_2)$ = $10 \, N$.
Volume $(V_2)$ = $10 \, mL$.
Equivalents of $H_2SO_4$ = $N_2 \times V_2 = 10 \times 10 = 100 \, meq$.
Since the equivalents of acid and base are equal $(100 = 100)$,the resulting solution is neutral.

(A) Let the weight of the partially dried soil be $100 \, g$.
In this soil,silica = $50 \, g$ and water = $7 \, g$.
Since only water evaporates,the amount of silica remains constant at $50 \, g$.
Let the weight of the original soil be $W \, g$.
In the original soil,water = $12 \%$ of $W = 0.12W$.
Therefore,the amount of non-water components (silica + others) = $W - 0.12W = 0.88W$.
Since the silica content is $50 \, g$ in the dried soil,and the non-water part of the original soil is $0.88W$,we equate the non-water mass:
$0.88W = (100 - 7) = 93 \, g$.
$W = 93 / 0.88 \approx 105.68 \, g$.
Now,the percentage of silica in the original soil is:
$\text{Percentage} = (\text{Mass of silica} / \text{Total mass of original soil}) \times 100$.
$\text{Percentage} = (50 / 105.68) \times 100 \approx 47.31 \%$.
Rounding to the nearest integer,we get $47 \%$.

(B) $20\%$ $H_2O_2$ by mass means $20 \ g$ of $H_2O_2$ in $100 \ g$ of solution.
Mass of $H_2O = 100 \ g - 20 \ g = 80 \ g$.
Moles of $H_2O_2 (n_1) = \frac{20}{34} \approx 0.588 \ mol$.
Moles of $H_2O (n_2) = \frac{80}{18} \approx 4.444 \ mol$.
Total moles $(n_{total}) = n_1 + n_2 = \frac{20}{34} + \frac{80}{18} = \frac{10}{17} + \frac{40}{9} = \frac{90 + 680}{153} = \frac{770}{153}$.
Mole fraction of water $(X_{H_2O}) = \frac{n_2}{n_{total}} = \frac{80/18}{770/153} = \frac{80}{18} \times \frac{153}{770} = \frac{40}{9} \times \frac{153}{770} = \frac{40 \times 17}{770} = \frac{680}{770} = \frac{68}{77}$.

(A) solution is neutral if the total positive charge equals the total negative charge or if the concentration of $H^{+}$ ions equals the concentration of $OH^{-}$ ions.
$(A)$ Total positive charge = $50 \times 2 = 100$. Total negative charge = $100 \times 1 = 100$. Since total positive charge equals total negative charge,the solution is neutral.
$(B)$ Total positive charge = $50 \times 2 = 100$. Total negative charge = $50 \times 1 = 50$. The solution is not neutral.
$(C)$ Total positive charge = $50 \times 1 = 50$. Total negative charge = $(25 \times 2) + (25 \times 1) = 75$. The solution is not neutral.
$(D)$ Total positive charge = $150 \times 3 = 450$. Total negative charge = $100 \times 3 = 300$. The solution is not neutral.

(C) Mass of $1 \, L$ of steam = Volume $\times$ Density = $1000 \, cm^3 \times 0.0006 \, g \cdot cm^{-3} = 0.6 \, g$.
Volume occupied by $0.6 \, g$ of liquid water = $\frac{\text{Mass}}{\text{Density}} = \frac{0.6 \, g}{1.0 \, g \cdot cm^{-3}} = 0.6 \, cm^3$.

(A) The white enamel of our teeth is composed of hydroxyapatite,which is a crystalline form of calcium phosphate,$Ca_{10}(PO_4)_6(OH)_2$. However,in the context of common chemistry curriculum questions regarding fluoridation,it is often associated with fluoroapatite,$3[Ca_3(PO_4)_2] \cdot CaF_2$. Among the given options,$Ca_3(PO_4)_2$ is the primary constituent of the enamel.

(A) The total number of equivalents in the mixture is given by the sum of equivalents of individual acids: $N_1V_1 + N_2V_2 + N_3V_3 = N_{resultant} \times V_{total}$.
Given:
$N_1V_1 = 50 \ mL \times 10 \ N = 500 \ meq$
$N_2V_2 = 25 \ mL \times 12 \ N = 300 \ meq$
$N_3V_3 = 40 \ mL \times 5 \ N = 200 \ meq$
Total equivalents = $500 + 300 + 200 = 1000 \ meq$.
Final volume of the mixture = $1000 \ mL$.
Normality of the resultant solution $(N)$ = $\frac{\text{Total equivalents}}{\text{Total volume in mL}} = \frac{1000 \ meq}{1000 \ mL} = 1 \ N$.

(D) To find the maximum mass,calculate the mass for each option:
$(A)$ $0.1 \ g$ atom of $C$ means $0.1 \ mole$ of $C$ atoms. Mass $= 0.1 \times 12 = 1.2 \ g$.
$(B)$ $0.1 \ mole$ of $NH_3$. Molar mass of $NH_3 = 14 + 3 = 17 \ g/mol$. Mass $= 0.1 \times 17 = 1.7 \ g$.
$(C)$ $6.02 \times 10^{22}$ molecules of $H_2$. Number of moles $= \frac{6.02 \times 10^{22}}{6.02 \times 10^{23}} = 0.1 \ mole$. Mass $= 0.1 \times 2 = 0.2 \ g$.
$(D)$ $1120 \ mL$ of $CO_2$ at $STP$ $(1 \ atm, 273 \ K)$. Number of moles $= \frac{1120}{22400} = 0.05 \ mole$. Mass $= 0.05 \times 44 = 2.2 \ g$.
Comparing the values: $1.2 \ g, 1.7 \ g, 0.2 \ g, 2.2 \ g$. The maximum mass is $2.2 \ g$.

(D) The thermal decomposition of sodium bicarbonate is given by: $2NaHCO_{3(s)} \xrightarrow{\Delta} Na_2CO_{3(s)} + H_2O_{(g)} + CO_{2(g)}$
When phenolphthalein is used as an indicator,the titration of $Na_2CO_3$ with $HCl$ proceeds only up to the formation of $NaHCO_3$:
$Na_2CO_3 + HCl \xrightarrow{Ph} NaHCO_3 + NaCl$
Moles of $HCl$ used $= 0.1 \ M \times 0.5 \ L = 0.05 \ mol$
From the stoichiometry of the titration reaction,$1 \ mol$ of $Na_2CO_3$ reacts with $1 \ mol$ of $HCl$. Thus,moles of $Na_2CO_3$ produced $= 0.05 \ mol$.
From the decomposition reaction,$2 \ mol$ of $NaHCO_3$ produce $1 \ mol$ of $Na_2CO_3$. Therefore,initial moles of $NaHCO_3 = 2 \times 0.05 \ mol = 0.1 \ mol$.
Mass of $NaHCO_3 = 0.1 \ mol \times 84 \ g/mol = 8.4 \ g$.
$PERCENT$ purity $= (8.4 \ g / 10 \ g) \times 100 = 84 \%$.

(A) Energy of one photon $E = \frac{12400 \ eV \mathring{A}}{3100 \mathring{A}} = 4 \ eV$.
Converting this energy to $kJ \ mol^{-1}$: $E = 4 \times 96 \ kJ \ mol^{-1} = 384 \ kJ \ mol^{-1}$.
The energy used to break the bond is $288 \ kJ \ mol^{-1}$.
The remaining energy is converted into $K.E.$: $K.E. = 384 \ kJ \ mol^{-1} - 288 \ kJ \ mol^{-1} = 96 \ kJ \ mol^{-1}$.
Percentage of energy converted to $K.E. = \frac{96}{384} \times 100 = 25 \%$.

(D) The molecular formula of urea is $(NH_2CONH_2)$.
To find the total number of electrons,we sum the atomic numbers of all atoms present in the molecule:
- $N$ (Nitrogen): Atomic number $= 7$,there are $2$ atoms: $2 \times 7 = 14$.
- $H$ (Hydrogen): Atomic number $= 1$,there are $4$ atoms: $4 \times 1 = 4$.
- $C$ (Carbon): Atomic number $= 6$,there is $1$ atom: $1 \times 6 = 6$.
- $O$ (Oxygen): Atomic number $= 8$,there is $1$ atom: $1 \times 8 = 8$.
Total electrons $= 14 + 4 + 6 + 8 = 32$.

(C) $A. 4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_{3(g)}$ (Gas $PH_3$ is evolved).
$B. 2PbO_2(s) \xrightarrow{\Delta} 2PbO(s) + O_{2(g)}$ (Gas $O_2$ is evolved).
$C. BaS(s) + ZnSO_4(aq.) \to ZnS(s) + BaSO_4(s)$ (No gas is evolved; both products are solids).
$D. S + 2H_2SO_4(conc.) \to 3SO_{2(g)} + 2H_2O(\ell)$ (Gas $SO_2$ is evolved).
Therefore,the correct option is $C$.

(A) The green compound $(A)$ is $FeSO_4 \cdot 7H_2O$. Upon heating, it decomposes to give $Fe_2O_3$ (red-brown residue), $SO_2$ (gas $(B)$), and $SO_3$ (gas $(C)$).
$SO_2$ reduces $KMnO_4$ (positive test).
$(C) = SO_3$.
$(C) + H_2SO_4 \to H_2S_2O_7$ (Oleum, $(D)$).
$(D) + [O] \to H_2S_2O_8$ (Peroxydisulphuric acid, $(E)$).
Structure of $H_2S_2O_8$ (Marshall's acid): $HO-SO_2-O-O-SO_2-OH$.
Counting $\sigma$ bonds in $H_2S_2O_8$: $2$ $(S-OH)$ bonds, $6$ $(S=O)$ bonds (each has $1 \sigma$), $1$ $(S-O-S)$ linkage ($2 \sigma$ bonds), and $1$ $(O-O)$ bond ($1 \sigma$ bond). Total $\sigma$ bonds = $2 + 6 + 2 + 1 = 11$.
$(F) = H_2O_2$, $(G) = K_4[Fe(CN)_6]$, $(H) = CO$.

(A) The density of water $(H_2O)$ is $1 \ g \ mL^{-1}$.
Given volume of $H_2O = 5.6 \ L = 5600 \ mL$.
Mass of $H_2O = \text{Density} \times \text{Volume} = 1 \ g \ mL^{-1} \times 5600 \ mL = 5600 \ g$.
Given $1 \ g$ atom of $Zn$ means $1 \ \text{mole}$ of $Zn$ atoms.
Atomic mass of $Zn = 65.5 \ g \ mol^{-1}$.
Mass of $Zn = 1 \ mol \times 65.5 \ g \ mol^{-1} = 65.5 \ g$.
Total mass of the mixture = $\text{Mass of } H_2O + \text{Mass of } Zn = 5600 \ g + 65.5 \ g = 5665.5 \ g$.

(A) $1$. Calculate moles of reactants: Moles of $N_2 = \frac{168 \ \text{g}}{28 \ \text{g/mol}} = 6 \ \text{mol}$. Moles of $H_2 = \frac{30 \ \text{g}}{2 \ \text{g/mol}} = 15 \ \text{mol}$.
$2$. Identify limiting reagent: According to the stoichiometry $N_2 + 3H_2 \rightarrow 2NH_3$,$1 \ \text{mol}$ of $N_2$ requires $3 \ \text{mol}$ of $H_2$. For $6 \ \text{mol}$ of $N_2$,we need $18 \ \text{mol}$ of $H_2$. Since we only have $15 \ \text{mol}$ of $H_2$,$H_2$ is the limiting reagent.
$3$. Calculate moles of $NH_3$ produced: $3 \ \text{mol}$ of $H_2$ produce $2 \ \text{mol}$ of $NH_3$. Thus,$15 \ \text{mol}$ of $H_2$ produce $\frac{2}{3} \times 15 = 10 \ \text{mol}$ of $NH_3$.
$4$. Calculate $\Delta n_g$: $\Delta n_g = \sum n_p - \sum n_r = 10 - (5 + 15) = -10 \ \text{mol}$ for the actual reaction occurring.
$5$. Calculate $\Delta U$: $\Delta U = \Delta H - \Delta n_g RT$. Given $\Delta H$ for $2 \ \text{mol}$ of $NH_3$ is $-24 \ \text{kcal}$,for $10 \ \text{mol}$ it is $-24 \times 5 = -120 \ \text{kcal}$.
$6$. $\Delta U = -120 - (-10 \times 0.002 \ \text{kcal/mol K} \times 700 \ \text{K}) = -120 + 14 = -106 \ \text{kcal}$.

(A) To find the order of increasing masses,we calculate the mass of each component:
$(I)$ Mass of $0.5$ mole of $O_3 = 0.5 \times 48 \, g = 24 \, g$.
$(II)$ $0.5 \, g$ atom of oxygen means $0.5$ mole of $O$ atoms. Mass $= 0.5 \times 16 \, g = 8 \, g$.
$(III)$ Number of moles of $O_2 = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} = 0.5$ mole. Mass $= 0.5 \times 32 \, g = 16 \, g$.
$(IV)$ Number of moles of $CO_2 = \frac{5.6 \, L}{22.4 \, L/mol} = 0.25$ mole. Mass $= 0.25 \times 44 \, g = 11 \, g$.
Comparing the masses: $8 \, g (II) < 11 \, g (IV) < 16 \, g (III) < 24 \, g (I)$.
Thus,the correct order is $II < IV < III < I$.

(C) Initial composition of $200 \ mL$ mixture:
$H_2 = 50\% \text{ of } 200 = 100 \ mL$
$CH_4 = 40\% \text{ of } 200 = 80 \ mL$
$CO_2 = 10\% \text{ of } 200 = 20 \ mL$
Combustion reactions:
$2H_2(g) + O_2(g) \to 2H_2O(l)$
$100 \ mL \ H_2$ consumes $50 \ mL \ O_2$ and produces $0 \ mL \ CO_2$.
$CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$
$80 \ mL \ CH_4$ consumes $160 \ mL \ O_2$ and produces $80 \ mL \ CO_2$.
Total $CO_2$ produced = $80 \ mL$ (from $CH_4$) + $20 \ mL$ (initial $CO_2$) = $100 \ mL$.
When passed through $aq. KOH$,$CO_2$ is absorbed:
$CO_2(g) + 2KOH(aq) \to K_2CO_3(aq) + H_2O(l)$
Volume contraction = Volume of $CO_2$ absorbed = $100 \ mL$.